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STatiana [176]
3 years ago
7

A 0.272-kg volleyball approaches a player horizontally with a speed of 12.6 m/s. The player strikes the ball with her fist and c

auses the ball to move in the opposite direction with a speed of 21.6 m/s. (a) What impulse is delivered to the ball by the player? (Take the direction of final velocity to be the positive direction. Indicate the direction with the sign of your answer.) kg · m/s (b) If the player's fist is in contact with the ball for 0.0600 s, find the magnitude of the average force exerted on the player's fist. N
Physics
1 answer:
Nady [450]3 years ago
6 0

(a) +9.30 kg m/s

The impulse exerted on an object is equal to its change in momentum:

I= \Delta p = m \Delta v = m (v-u)

where

m is the mass of the object

\Delta v is the change in velocity of the object, with

v = final velocity

u = initial velocity

For the volleyball in this problem:

m = 0.272 kg

u = -12.6 m/s

v = +21.6 m/s

So the impulse is

I=(0.272 kg)(21.6 m/s - (-12.6 m/s)=+9.30 kg m/s

(b) 155 N

The impulse can also be rewritten as

I=F \Delta t

where

F is the force exerted on the volleyball (which is equal and opposite to the force exerted by the volleyball on the fist of the player, according to Newton's third law)

\Delta t is the duration of the collision

In this situation, we have

\Delta t = 0.06 s

So we can re-arrange the equation to find the magnitude of the average force:

F=\frac{I}{\Delta t}=\frac{9.30 kg m/s}{0.06 s}=155 N

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1. You place an object 63 cm in front of a converging lens, with a 40 cm focal length.
NikAS [45]

Answer:

1.

109.6 cm ,  - 1.74 , real

2.

1.5

Explanation:

1.

d₀ = object distance = 63 cm

f = focal length of the lens = 40 cm

d = image distance = ?

using the lens equation

\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}

\frac{1}{40} = \frac{1}{63} + \frac{1}{d}

d = 109.6 cm

magnification is given as

m = \frac{-d}{d_{o}}

m = \frac{-109.6}{63}

m = - 1.74

The image is real

2

d₀ = object distance = a

d = image distance = - (a + 5)

f = focal length of lens = 30 cm

using the lens equation

\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}

\frac{1}{30} = \frac{1}{a} + \frac{1}{- (a + 5)}

a = 10

magnification is given as

m = \frac{-d}{d_{o}}

m = \frac{- (- (a +5))}{a}

m = \frac{(5 + 10)}{10}

m = 1.5

6 0
2 years ago
Which statement is the correct representation of these electric field lines?
GuDViN [60]
I would say a. hope this helps
4 0
3 years ago
Read 2 more answers
The low-frequency speaker of a stereo set has a surface area of 0.06 m2 and produces 2.03 W of acoustical power. What is the int
zimovet [89]

Answer:

33.83W/m²

Explanation:

The intensity of the speake at the surface is

I = P/A

I = 2.03W / 0.06m²

I = 33.83W/m²

8 0
2 years ago
interval (from lowering the body to his feet taking off) lasts for 0.3 s and the mass of the player is 90 kg. Ignore air resista
steposvetlana [31]

Answer:

The force applied to the surface is 9 kilo Newton.

Explanation:

While jumping on the surface the player applies the force that is equal to its weight on the surface.

The mass of the player is given as 90 kg.

Force applied by the player = weight of the player

Force applied by the player = m × g

Where m is the mass of the player and g is acceleration due to gravity

Force applied by the player = 90 × 9.8

Force applied by the player = 882 Newton

Expressing your answer to one significant figure, we get

Force applied by the player =0. 9 kilo Newton

The force applied to the surface is 0.9 kilo Newton.

8 0
3 years ago
PLEASE HELP URGENT: Please see attachment for problem! AP physics projectile question. I'd REALLY appreciate any help.
kiruha [24]
I’m so sorry, I need more information. Good luck and I’m sorry I couldn’t help you :(
4 0
3 years ago
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